UP TGT Mathematics Mock Test - 1 - UPTET MCQ

• Pair 1 is correctly matched: Non-violence (Ahimsa) is a core principle of Jainism, strongly associated with Mahavira.
• Pair 2 is correctly matched: The Eightfold Path is a fundamental teaching of Buddhism, introduced by Gautama Buddha as a means to end suffering.
• Pair 3 is incorrectly matched: Discarding clothes is associated with Mahavira and the Digambara sect of Jainism, not Chandragupta Maurya.

• The correct answer is Guru Nanak is the only Guru.

Key Points

• The Beliefs and Faiths of the Kuka Sect:
  • The sect believes that Adi Granthis the only true holy book of their religion.
  • Gobind Singh is the only Guru.
  • Any person, irrespective of caste or religion, can be admitted as a Namdhari convert.
  • Sodhis, Bedis, Mahants, Brahmins and such like are impostors, as none are Gurus except Gobind Singh.
  • It’s worth note that among Sikhs the Sodhis and Bedis had started getting worshipped during those times.
  • Devidwaras, Shivdwaras and Mandirs are a means of extortion, to be held in contempt and never visited.
  • Idols and idol-worship are insulting to God, and will not be forgiven. The Namdharis were iconoclasts.
  • Converts are allowed to read Gobind Singh’s Grantha and no other book.
  • Pure vegetarianism. It was against killing of cattle and kine.
  • No caste system

The correct answer is Khizr Khan.

• Khizr Khan was the governor of Multan and Timur's deputy in India..
• Khizr Khan was the founder of the Sayyid Dynasty.
• This dynasty governed for 37 years from 1414 to 1451 AD by four rulers- Khizr Khan, Mubarak, Muhammad Shah, Alam Shah.
• About Khizr Khan(1414-1421):
  • He was the founder of the Sayyid Dynasty in India and governed as a deputy of Timur’s son and heir, Shah Rukh.
  • His reign was regarded as utter chaos and disorder.
  • The empire’s territory had Shrunken to Delhi and adjacent areas and even these parts were frequently challenged by the Hindu Zamindars of Etawah, Katehar, Kannauj, Patiala and Kampila.
  • He died of illness in AD 1421.

• It was started by Sir Syed Ahmad Khan (1817-1898) for the Muslims' social and educational advancement in India.
• He fought against the medieval backwardness and advocated a rational approach towards religion. In 1866, he started the Muhammadan Educational Conference as a general forum for spreading liberal ideas among the Muslims.
• In 1875, he founded a modem school at Aligarh to promote English education among the Muslims. This had later grown into the Muhammadan Anglo-Oriental College and then into the Aligarh Muslim University.

The correct answer is Lala Lajpat Rai.

Key Points

• The Swadeshi movement united the dispersed leadership of India, which awakened the entire section of society such as Women, students, and a large section of the urban and rural population into active politics for the first time.
• Here is the list of personalities associated with the Swadeshi Movement of British India for general awareness about the personalities who shape the true nature of India’s Freedom Struggle:
• Lala Lajpat Rai- He took the movement to Punjab and northern India.
• Syed Haider Raza- He popularised the Swadeshi Movement in Delhi.
• Bal Gangadhar Tilak- He spread the message of swadeshi to Poona and Bombay, and organized Ganapati and Shivaji festivals to arouse patriotic feelings. He stressed that the aim of swadeshi, boycott, and national education was the attainment of Swaraj. He opened cooperative stores and headed the Swadeshi Vastu Pracharini Sabha.

Girija Subramanian, the current Chairperson and MD of Agriculture Insurance Company (AIC), has been selected as the next CMD of New India Assurance by the Financial Services Institutions Bureau (FSIB). This selection process also included appointments for other executive positions in PSU general insurers.

Explanation:

If A is the arithmetic mean between a and b, 

If G is the geometric mean between a and b,

⇒ G = √ab

If H is the harmonic mean between a and b,

Now, AH = 

⇒ AH = ab

⇒ AH = G²

This is a form of a G.P.

Hence, If the arithmetic mean, geometric mean, and Harmonic mean between two numbers 'a' and 'b' are A, G, and H respectively, then A, G, H will be in geometric series.

Concept:

1. A complex number (Z):  Complex number is the combination of a real number and an imaginary number. It is given by

Z = x + iy, where 'x' and 'y' are the real and imaginary parts of Z and

i = √-1 or i² = -1

Re(Z) = x and Img(Z) = y

2. Two complex numbers will be equal if their real and imaginary part are equal.

Calculation:

Given that,

⇒(1 + i)(3x - ix) - 2i(3 - i) + (2 - 3i)(3y + iy) + i(3 + i) = 10i

⇒ 3x + 3xi - ix - i²x - 6i + 2i² + 6y - 9iy + 2iy - 3yi² + 3i + i² = 10i

We know that,  i² = -1

⇒ 3x + 2xi + x - 6i - 2 + 6y - 7yi + 3y + 3i -1 = 10i  

⇒ 4x + 9y − 3 + 2xi − 7yi − 13i = 0

⇒ 4x + 9y − 3 + (2x − 7y − 13)i = 0

On comparing the real part and imaginary part, we get

4x + 9y − 3 = 0     …… (1)

2x − 7y − 13 = 0  …… (2)

On solving both equations, we get

x = 3 and y = −1

Hence, the value of x, y is 3, −1.

Concept:

Sphere:

General equation of sphere is given as x² + y² + z² + 2ux + 2vy + 2wz + d = 0

• Centre of the sphere is (-u, -v, -w)
• Radius = 

Standard equation of sphere is given as (x - h)² + (y - k)² + (z - p)² = r²

• Centre of the sphere is (h, k, p)
• Radius = r

Calculation:

Given:

Centre = (0, 0, 0)

Radius = r = 1

Now, the equation of a sphere is (x - 0)² + (y - 0)² + (z - 0)² = 1²

∴ the equation of a sphere is x² + y² + z² = 1

Concept:

• →FIf the two vectors are    and  then the dot created is articulated as 
• Let’s suppose these two vectors are separated by angle θ. 
• To know what’s the angle measurement we solve with the below form

The dot product is given by:

Thus, the angle between two vectors formula is given by,

where θ is the angle between    and 
  

Concept:

Square matrix A is said to be skew-symmetric if a_ij = −a_ij for all i and j.

Square matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A ⇔ A^T = −A

All the main diagonal elements in the skew-symmetric matrix are zero.

Calculation:

For a skew-symmetric matrix, diagonal elements are zero and AT = −A

So, both and are Skew-symmetric matrix.

Concept:

If the differential equation is in the form  where f(x, y) and ϕ(x, y) are homogeneous functions of the same degree in x and y

To solve this homogeneous equation

(i) Put y = vx, then 

(ii) Separate the variables v and x and integrate

Calculation:

The given differential equation is

(x + y) (dx – dy) = dx + dy

⇒ (x + y) dx – (x + y) dy = dx + dy

⇒ (x + y – 1) dx = (x + y + 1) dt

Let,  

Concept:

when z = x + iy then, Principal amplitude of a complex number, θ = tan^-1()

tan  = 

x > 0, y < 0, The point lies in IVth quadrant.

Calculation:

Let θ be the principal value of amplitude of   - i

Since, tan θ =  and - i lies in IV^th quadrant.

 tan θ = tan (-), θ = -

Calculation:

Given, f(x) = 2x, g(x) = x² + 2

then, (f + g)(2) = f(2) + g(2)

= (2 × 2) + (2² + 2)

= 4 + 6

= 10

Explanation:

Given series is,

By ratio test, the given series converges for |x| < 1 and diverges for |x| > 1

Let us examine the series for x = ± 1

For x = 1, the series reduces to

This is an alternating series and is convergent.

For x = -1 the series becomes

This is a divergent series as can be seen by comparison with P-series with P = 1

Hence the given series is converges for -1 < x < 1

Formula Used:

Sum of square of n numbers =  

Calculation:   

Case: 1

12 + 22 + 32 + . . . . . . + 202

Sum of square of 20 numbers = 

⇒ Sum of square of 20 numbers = 

⇒ Sum of square of 20 numbers = 2870

Case: 2

 12 + 22 + 32 + . . . . . . + 102

⇒ Sum of square of 10 numbers = $\frac{10(10+1)(2\times 10+1)}{6}$

⇒ Sum of square of 10 numbers = 

⇒ Sum of square of 10 numbers = 385

According to the question: 

⇒ 112 + 122 + 132 + . . . . . . + 202 = Sum of square of 20 numbers - Sum of square of 10 numbers 

⇒ 112 + 122 + 132 + . . . . . . + 202 = 2870 - 385 

∴ 112 + 122 + 132 + . . . . . . + 202 = 2485

Concept:

• sin (A - B) = sin A cos B - cos A sin B
• The maximum value of a sin θ + b cos θ is 

Calculation:

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

Apply C2 → C2 + C3 

Taking common (a + b + c) from column 2, we get

As we can see that the first and the second column of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

Concept:

If f(x) =  such that, 

,

{ form }

then, by L's hospital Rule,

Calculation:

∴ The correct answer is option (4).

Formula used:

• sin θ/cos θ = tan θ 
• cos θ/sin θ = cot θ 
• sin2θ + cos2θ = 1
• 2sin θ cos θ = sin 2θ

Calculation:

∴  f(x)_max = 

CONCEPT:

Equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2

CALCULATION:

Here, we have to find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4

First let's find the point of intersection of the lines  3x + 2y = 11 and 2x + 3y = 4

So, by solving the equations  3x + 2y = 11 and 2x + 3y = 4, we get x = 5 and y = - 2

So, the required circle passes through the point (5, - 2)

Let the radius of the required circle be r

As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2

Here, we have h = 2 and k = - 3

⇒ (x - 2)² + (y + 3)² = r² ------------(1)

∵ The circle  passes through the point (5, - 2)

So, x = 5 and y = - 2 will satisfy the equation (1)

⇒ (5 - 2)² + (- 2 + 3)² = r²

⇒ r² = 10

So, the equation of the required circle is (x - 2)2 + (y + 3)2 = 10

⇒ x² + y² - 4x + 6y + 3 = 0

So, the equation of the required circle is x2 + y2 - 4x + 6y + 3 = 0

Hence, option A is the correct answer.

Concept:

If 1, ω, ω² are the cube roots of unity,

•  and 
•       ----(1)

• 1 + ω + ω2 = 0 and ω³ = 1      ----(2)

Calculation:

We have (1 + ω2)(1 + ω4)(1 + ω8)(1 + ω16)

⇒ (1 + ω² + ω⁴ + ω⁶)(1 + ω¹⁶ + ω⁸ + ω²⁴)

⇒ (1 + ω2 + ω + 1)(1 + ω + ω² + 1)      [using (1)]

⇒ (0 + 1)(0 + 1)      [using (2)]

⇒ 1

Hence, (1 + ω2)(1 + ω4)(1 + ω8)(1 + ω16) = 1

CONCEPT:

• Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
  • Initial Velocity: The initial velocity can be given as x components and y components.

u_x = u cosθ

u_y = u sinθ

Where u stands for initial velocity magnitude and θ refers to projectile angle.

• Maximum Height: The maximum height is reached when v_y = 0.

Where h is the maximum height.

CALCULATION:

Given that u_x = 60 m/s, θ = 60°

At maximum height, there is only a horizontal velocity component and it is always constant because there is no acceleration in the horizontal direction.

u_x = u cosθ → horizontal component at maximum height.

60 = u cos60° ⇒ u = 120 m/s .

Concept:

If for each x ∈ A there exist only one image y ∈ B and each y ∈ B has a unique pre-image x ∈ A (i.e. no two elements of A have the same image in B), then f is said to be a one-one function. 

The mapping of 'f' is said to be onto if every element of Y is the f-image of at least one element of X.

​   

A function f from A (the domain) to B (the range) is one-to-one and onto when no element of B is the image of more than one element in A, and all elements in B are used.

Calculation:

f (x) = x⁴ 

first we check for one - one,

f (x₁) = ( x₁)⁴

f (x₂) = ( x₂)⁴ 

Putting , f (x₁) = f (x₂)

⇒ (x₁)⁴ = (x₂)⁴ 

⇒ x₁ = x₂  or  x₁ = -x₂ 

Thus , x₁ doest not have unique image , it  has two image , x₂ and - x₂  . 

Thus, it is not a one-one function. 

Check for onto ,​  

let , f ( x ) = y , where y ϵ R 

x⁴ = y 

⇒ x = ± y^1/4 

y is real number , it can be negative or positive . 

Hence x is not real 

∴ f is not onto 

Hence, given function f is neither one-one nor onto.

The correct option is 2.

Given:

AB = AC and ∠BAC = 48°

Concept used:

The sum of the opposite angles of a cyclic quadrilateral = 180°

Calculation:

∠ABC = ∠ACB  [∵ AB = AC]

∠BAC + ∠ABC + ∠ACB = 180°

So, ∠ABC = ∠ACB = (180° - 48°)/2 = 132°/2 = 66°

Also, ∠ADC + ∠ABC = 180°

⇒ ∠ADC + 66° = 180°

⇒ ∠ADC = 180° - 66° = 114°

∴ The measure of ∠ADC is 114°

Given:

A = 

Concept used:

Transpose of the Matrix: Interchange of Row into Column. 

or Interchange of Column into Row.

Calculations:

A = 

⇒ A^t  = B = 

∴ a₂₂ = -90 (The entry which lies in 2nd row and 2nd column)

∴ option 3 is correct 

Concept:

The general second degree equation of a circle in x and y is given by: a ⋅ x2 + a ⋅ y2 + 2gx + 2fy + c = 0 with centre (-g, -f) and radius .

Let C (h, k) be the centre of the circle and r be the radius of the circle, then the equation of the circle in the standard form is given by: 

Let us consider two circles represented by the equations shown below: S1: x2 + y2 + 2g1x + 2f1y + c1 = 0 and S2: x2 + y2 + 2g2x + 2f2y + c2 = 0.Then the equation of common chord to the above two circles is given by: S1 – S2 = 0 or 2 ⋅ (g1 – g2) ⋅ x + 2 ⋅ (f1 – f2) ⋅ y + 2 ⋅ (c1 – c2) = 0.

If C1C2 = |r1 – r2| then given two circles touch each other internally. Hence, number of common tangents to the given circles in this case = 1.

Calculation:

Given: S1 = x2 + y2 = 4 and S2 = x2 + y2 - 6x - 8y - 24 = 0.

Let  C1 and C2 be the centres and r1 and r2 be the radius of the circles S1 and S2 respectively.

Now by comparing equation of circle S1 with , we get

⇒ C1 = (0, 0) and r1 = 2

Similarly by comparing the equation of circle S2 with a ⋅ x2 + a ⋅ y2 + 2gx + 2fy + c = 0 , we get a = 1, g = - 3, f = - 4 and c = - 24

⇒ C2 = (3, 4) and radius r2 = 7.

∴ C1C2 = 5 and |r1 - r2| = 5

⇒ C1C2 = |r1 - r2|

As we know that, if C1C2 = |r1 – r2| then given two circles touch each other internally. Hence, number of common tangents to the given circles in this case = 1.

Hence, option A is true.

Concept:

Parabola:

• The focus of the parabola y² = 4ax is at (a, 0).
• The latus rectum of the parabola y² = 4ax cuts the parabola at (a, 2a) and (a, -2a).

Area under a curve:

• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral , for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.

Calculation:

Since the graph of the parabola y² = 4ax is symmetrical about the x-axis, the required area is:

2 ×  

= 2 × 2√a  

= 2 × 2√a [23x32]0a 

=  = .

Additional Information
The latus rectum is a line which passes through the focus and is parallel to the directrix. 

Concept:

The general equation of the ellipse is:

Here, coordinates of foci are (±ae, 0).

Also, we have b² = a²(1 - e²), where e is the eccentricity.

Calculation:

Since the coordinates of the foci are (±3, 0).

⇒ ae = 3

⇒ a × (1/3) = 3      (∵ e = 1/3)

⇒ a = 9

Now, b2 = a2(1 - e2)

$\Rightarrow b^2=81\left(1-\frac{1}{9}\right)$

⇒ b² = 72

On putting the value of a² and b² in the general equation of an ellipse, we get

Hence, the equation of the ellipse is .

Concept Used:

1. a logx = log (x^a)

2. e^log(n)  = n

3. ∫xⁿdx = 

Application:

We have,

I = e5 log x

or, I =  = x⁵

Hence, ∫x⁵ dx = x⁶/6 + C